NCERT Solutions for Class 7 Maths Chapter 10 Algebraic Expressions Ex 10.2

Exercise 10.2

1. If $\mathbf{m = 2}$, find the value of:

(i)$\mathbf{m - 2}$

Ans: Substitute $2$ for $m$ and solve.

$ \Rightarrow m - 2 = 2 - 2 \\ $

$ \Rightarrow m - 2 = 0  \\ $

The value is $0$.

(ii) $\mathbf{3m - 5}$

Ans: Substitute $2$ for $m$ and solve.

$ \Rightarrow 3m - 5 = 3\left( 2 \right) - 5 \\$

$ \Rightarrow 3m - 5 = 6 - 5 \\ $

$ \Rightarrow 3m - 5 = 1 \\ $

The value is $1$.

(iii) $\mathbf{9 - 5m}$

Ans: Substitute $2$ for $m$ and solve.

$ \Rightarrow 9 - 5m = 9 - 5\left( 2 \right) \\ $

$ \Rightarrow 9 - 5m = 9 - 10  \\ $

$ \Rightarrow 9 - 5m =  - 1 \\ $

The value is $ - 1$.

(iv)$\mathbf{3{m^2} - 2m - 7}$

Ans: Substitute $2$ for $m$ and solve.

$ \Rightarrow 3{m^2} - 2m - 7 = 3{\left( 2 \right)^2} - 2\left( 2 \right) - 7 \\ $

$ \Rightarrow 3{m^2} - 2m - 7 = 3\left( 4 \right) - 4 - 7  \\ $

 $ \Rightarrow 3{m^2} - 2m - 7 = 12 - 11  \\ $

  $ \Rightarrow 3{m^2} - 2m - 7 = 1 \\  $

The value is $1$.

(v) $\mathbf{\dfrac{{5m}}{2} - 4}$

Ans: Substitute $2$ for $m$ and solve.

$ \Rightarrow \dfrac{{5m}}{2} - 4 = \dfrac{{5 \times 2}}{2} - 4 \\ $

 $ \Rightarrow \dfrac{{5m}}{2} - 4 = 5 - 4 \\ $

 $ \Rightarrow \dfrac{{5m}}{2} - 4 = 1 \\ $

The value is $1$.

2. If $\mathbf{p =  - 2}$, find the value of:

(i) $\mathbf{4p + 7}$

Ans: Substitute $ - 2$ for $p$ and solve.

$ \Rightarrow 4p + 7 = 4\left( { - 2} \right) + 7 \\ $

$ \Rightarrow 4p + 7 =  - 8 + 7 \\ $

 $ \Rightarrow 4p + 7 =  - 1 \\ $

 The value is $ - 1$.

(ii) $\mathbf{ - 3{p^2} + 4p + 7}$

Ans: Substitute $ - 2$ for $p$ and solve.

$\Rightarrow  - 3{p^2} + 4p + 7 =  - 3{\left( { - 2} \right)^2} + 4\left( { - 2} \right) + 7 \\ $

  $ \Rightarrow  - 3{p^2} + 4p + 7 =  - 3\left( 4 \right) - 8 + 7 \\ $

  $ \Rightarrow  - 3{p^2} + 4p + 7 =  - 12 - 1 \\ $

  $ \Rightarrow  - 3{p^2} + 4p + 7 =  - 13 \\ $

The value is $ - 13$.

(iii) $\mathbf{ - 2{p^3} - 3{p^2} + 4p + 7}$

Ans: Substitute $ - 2$ for $p$ and solve.

$ \Rightarrow  - 2{p^3} - 3{p^2} + 4p + 7 =  - 2{\left( { - 2} \right)^3} - 3{\left( { - 2} \right)^2} + 4\left( { - 2} \right) + 7 \\ $

 $ \Rightarrow  - 2{p^3} - 3{p^2} + 4p + 7 =  - 2 \times \left( { - 8} \right) - 3\left( 4 \right) - 8 + 7 \\ $

$ \Rightarrow  - 2{p^3} - 3{p^2} + 4p + 7 = 16 - 12 - 1 \\ $

$ \Rightarrow  - 2{p^3} - 3{p^2} + 4p + 7 = 4 - 1 \\ $

$ \Rightarrow  - 2{p^3} - 3{p^2} + 4p + 7 = 3  \\ $

The value is $3$.

3. Find the value of the following expressions, when $\mathbf{x =  - 1}$:

(i) $\mathbf{2x - 7}$

Ans: Substitute $ - 1$ for $x$ and solve.

$ \Rightarrow 2x - 7 = 2\left( { - 1} \right) - 7 \\ $

$ \Rightarrow 2x - 7 =  - 2 - 7 \\ $

$ \Rightarrow 2x - 7 =  - 9 \\ $

The value of expression $2x - 7$ is $ - 9$.

(ii) $ \mathbf{- x + 2}$

Ans: Substitute $ - 1$ for $x$ and solve.

$ \Rightarrow  - x + 2 =  - \left( { - 1} \right) + 2  \\ $

$ \Rightarrow  - x + 2 = 1 + 2 \\ $

 $ \Rightarrow  - x + 2 = 3  \\ $ 

The value of expression $ - x + 2$ is $3$.

(iii) $\mathbf{{x^2} + 2x + 1}$

Ans: Substitute $ - 1$ for $x$ and solve.

$ \Rightarrow {x^2} + 2x + 1 = {\left( { - 1} \right)^2} + 2\left( { - 1} \right) + 1 \\ $

$ \Rightarrow {x^2} + 2x + 1 = 1 - 2 + 1  \\ $

$ \Rightarrow {x^2} + 2x + 1 =  - 1 + 1  \\ $

$ \Rightarrow {x^2} + 2x + 1 = 0 \\ $

The value of expression ${x^2} + 2x + 1$ is $0$.

(iv)$\mathbf{2{x^2} - x - 2}$

Ans: Substitute $ - 1$ for $x$ and solve.

$ \Rightarrow 2{x^2} - x - 2 = 2{\left( { - 1} \right)^2} - \left( { - 1} \right) - 2 \\ $

$ \Rightarrow 2{x^2} - x - 2 = 2 \times 1 + 1 - 2 \\ $

$ \Rightarrow 2{x^2} - x - 2 = 2 + 1 - 2 \\ $

$ \Rightarrow 2{x^2} - x - 2 = 3 - 2 \\ $

$ \Rightarrow 2{x^2} - x - 2 = 1 \\ $

The value of expression $2{x^2} - x - 2$ is $1$.

4. If $\mathbf{a = 2$, $b =  - 2}$, find the value of:

(i) $\mathbf{{a^2} + {b^2}}$

Ans: Substitute $2$ for $a$ and $ - 2$ for $b$ then solve.

$ \Rightarrow {a^2} + {b^2} = {\left( 2 \right)^2} + {\left( { - 2} \right)^2} \\ $

$ \Rightarrow {a^2} + {b^2} = 4 + 4  \\ $

$ \Rightarrow {a^2} + {b^2} = 8 \\ $

The value is $8$.

(ii) ${a^2} + ab + {b^2}$

Ans: Substitute $2$ for $a$ and $ - 2$ for $b$ then solve.

$ \Rightarrow {a^2} + ab + {b^2} = {\left( 2 \right)^2} + \left( 2 \right)\left( { - 2} \right) + {\left( { - 2} \right)^2} \\ $

$ \Rightarrow {a^2} + ab + {b^2} = 4 - 4 + 4 \\ $

$ \Rightarrow {a^2} + ab + {b^2} = 4 \\ $

The value is $4$.

(iii) ${a^2} - {b^2}$

Ans: Substitute $2$ for $a$ and $ - 2$ for $b$ then solve.

$ \Rightarrow {a^2} - {b^2} = {\left( 2 \right)^2} - {\left( { - 2} \right)^2}  \\ $

$ \Rightarrow {a^2} - {b^2} = 4 - 4 \\ $

$ \Rightarrow {a^2} - {b^2} = 0  \\ $

The value is $0$.

5. When $\mathbf{a = 0$, $b =  - 1}$, find the value of the given expressions:

(i) $\mathbf{2a + 2b}$

Ans: Substitute $0$ for $a$ and $ - 1$ for $b$ then solve.

$ \Rightarrow 2a + 2b = 2\left( 0 \right) + 2\left( { - 1} \right)  \\ $

$ \Rightarrow 2a + 2b = 0 - 2  \\ $

$ \Rightarrow 2a + 2b =  - 2  \\ $

The value of expression $2a + 2b$ is $ - 2$.

(ii) $\mathbf{2{a^2} + {b^2} + 1}$

Ans: Substitute $0$ for $a$ and $ - 1$ for $b$ then solve.

$ \Rightarrow 2{a^2} + {b^2} + 1 = 2{\left( 0 \right)^2} + {\left( { - 1} \right)^2} + 1  \\ $

$ \Rightarrow 2{a^2} + {b^2} + 1 = 2 \times 0 + 1 + 1 \\ $

$ \Rightarrow 2{a^2} + {b^2} + 1 = 0 + 2 \\ $

$ \Rightarrow 2{a^2} + {b^2} + 1 = 2 \\ $

The value of expression $2{a^2} + {b^2} + 1$ is $2$.

(iii) $\mathbf{2{a^2}b + 2a{b^2} + ab}$

Ans: Substitute $0$ for $a$ and $ - 1$ for $b$ then solve.

$ \Rightarrow 2{a^2}b + 2a{b^2} + ab = 2{\left( 0 \right)^2}\left( { - 1} \right) + 2\left( 0 \right){\left( { - 1} \right)^2} + \left( 0 \right)\left( { - 1} \right) \\ $

$ \Rightarrow 2{a^2}b + 2a{b^2} + ab = 0 + 0 + 0 \\ $

$ \Rightarrow 2{a^2}b + 2a{b^2} + ab = 0 \\ $

The value of expression $2{a^2}b + 2a{b^2} + ab$ is $0$.

(iv)$\mathbf{{a^2} + ab + 2}$

Ans: Substitute $0$ for $a$ and $ - 1$ for $b$ then solve.

$ \Rightarrow {a^2} + ab + 2 = {\left( 0 \right)^2} + \left( 0 \right)\left( { - 1} \right) + 2 \\$

$ \Rightarrow {a^2} + ab + 2 = 0 + 0 + 2 \\ $

$ \Rightarrow {a^2} + ab + 2 = 2 \\ $

The value of expression ${a^2} + ab + 2$ is $2$.

6. Simplify the expressions and find the value if $\mathbf{x}$ is equal to $\mathbf{2}$:

(i) $\mathbf{x + 7 + 4\left( {x - 5} \right)}$

Ans: Simplify the brackets and combine the like terms.

$ \Rightarrow x + 7 + 4\left( {x - 5} \right) = x + 7 + 4x - 20  \\ $

 $ \Rightarrow x + 7 + 4\left( {x - 5} \right) = x + 4x + 7 - 20 \\ $

 $ \Rightarrow x + 7 + 4\left( {x - 5} \right) = 5x - 13 \\ $

Substitute $2$ for $x$ and then solve.

$ \Rightarrow x + 7 + 4\left( {x - 5} \right) = 5\left( 2 \right) - 13  \\ $

$ \Rightarrow x + 7 + 4\left( {x - 5} \right) = 10 - 13 \\ $

$  \Rightarrow x + 7 + 4\left( {x - 5} \right) =  - 3 \\  $

 The determined value is $ - 3$.

(ii) $\mathbf{3\left( {x + 2} \right) + 5x - 7}$

Ans: Simplify the brackets and combine the like terms.

$ \Rightarrow 3\left( {x + 2} \right) + 5x - 7 = 3x + 6 + 5x - 7 \\ $

$ \Rightarrow 3\left( {x + 2} \right) + 5x - 7 = 3x + 5x + 6 - 7 \\ $

$ \Rightarrow 3\left( {x + 2} \right) + 5x - 7 = 8x - 1 \\  $

Substitute $2$ for $x$ and then solve.

$ \Rightarrow 3\left( {x + 2} \right) + 5x - 7 = 8\left( 2 \right) - 1 \\ $

$ \Rightarrow 3\left( {x + 2} \right) + 5x - 7 = 16 - 1  \\ $

$ \Rightarrow 3\left( {x + 2} \right) + 5x - 7 = 15 \\ $

The determined value is $15$.

(iii) $\mathbf{6x + 5\left( {x - 2} \right)}$

Ans: Simplify the brackets and combine the like terms.

$ \Rightarrow 6x + 5\left( {x - 2} \right) = 6x + 5x - 10 \\ $

$ \Rightarrow 6x + 5\left( {x - 2} \right) = 11x - 10  \\ $

Substitute $2$ for $x$ and then solve.

$ \Rightarrow 6x + 5\left( {x - 2} \right) = 11 \times 2 - 10  \\ $

$ \Rightarrow 6x + 5\left( {x - 2} \right) = 22 - 10   \\ $

$ \Rightarrow 6x + 5\left( {x - 2} \right) = 12  \\  $

The determined value is $12$.

(iv)$\mathbf{4\left( {2x - 1} \right) + 3x + 11}$

Ans: Simplify the brackets and combine the like terms.

$ \Rightarrow 4\left( {2x - 1} \right) + 3x + 11 = 8x - 4 + 3x + 11  \\ $

$ \Rightarrow 4\left( {2x - 1} \right) + 3x + 11 = 8x + 3x - 4 + 11  \\ $

$ \Rightarrow 4\left( {2x - 1} \right) + 3x + 11 = 11x + 7   \\ $

Substitute $2$ for $x$ and then solve.

$ \Rightarrow 4\left( {2x - 1} \right) + 3x + 11 = 11 \times 2 + 7 \\ $

$ \Rightarrow 4\left( {2x - 1} \right) + 3x + 11 = 22 + 7  \\ $

$ \Rightarrow 4\left( {2x - 1} \right) + 3x + 11 = 29  \\ $

The determined value is $29$.

7. Simplify these expressions and find their values if $\mathbf{x = 3$, $a =  - 1$, $b =  - 2}$:

(i) $\mathbf{3x - 5 - x + 9}$

Ans: Simplify the brackets and combine the like terms.

$ \Rightarrow 3x - 5 - x + 9 = 3x - x - 5 + 9 \\ $

$ \Rightarrow 3x - 5 - x + 9 = 2x + 4 \\ $

Substitute $3$ for $x$ and then solve.

$ \Rightarrow 3x - 5 - x + 9 = 2 \times 3 + 4 \\ $

$ \Rightarrow 3x - 5 - x + 9 = 6 + 4 \\ $

$ \Rightarrow 3x - 5 - x + 9 = 10 \\ $

The determined value is $10$.

(ii) $\mathbf{2 - 8x + 4x + 4}$

Ans: Simplify the brackets and combine the like terms.

$ \Rightarrow 2 - 8x + 4x + 4 =  - 8x + 4x + 2 + 4 \\ $

$ \Rightarrow 2 - 8x + 4x + 4 =  - 4x + 6  \\ $

Substitute $3$ for $x$ and then solve.

$ \Rightarrow 2 - 8x + 4x + 4 =  - 4 \times 3 + 6 \\ $

$ \Rightarrow 2 - 8x + 4x + 4 =  - 12 + 6 \\ $

$ \Rightarrow 2 - 8x + 4x + 4 =  - 6 \\ $

The determined value is $ - 6$.

(iii) $\mathbf{3a + 5 - 8a + 1}$

Ans: Simplify the brackets and combine the like terms.

$ \Rightarrow 3a + 5 - 8a + 1 = 3a - 8a + 5 + 1 \\ $

 $ \Rightarrow 3a + 5 - 8a + 1 =  - 5a + 6 \\ $

Substitute $ - 1$ for $a$ and then solve.

$ \Rightarrow 3a + 5 - 8a + 1 =  - 5\left( { - 1} \right) + 6 \\ $

$ \Rightarrow 3a + 5 - 8a + 1 = 5 + 6 \\ $

$ \Rightarrow 3a + 5 - 8a + 1 = 11 \\ $

The determined value is $11$.

(iv)$\mathbf{10 - 3b - 4 - 5b}$

Ans: Simplify the brackets and combine the like terms.

$ \Rightarrow 10 - 3b - 4 - 5b =  - 3b - 5b + 10 - 4 \\ $

$ \Rightarrow 10 - 3b - 4 - 5b =  - 8b + 6 \\  $

Substitute $ - 2$ for $b$ and then solve.

$ \Rightarrow 10 - 3b - 4 - 5b =  - 8\left( { - 2} \right) + 6  \\ $

$ \Rightarrow 10 - 3b - 4 - 5b = 16 + 6 \\ $

$ \Rightarrow 10 - 3b - 4 - 5b = 22 \\ $

The determined value is $22$.

(v) $\mathbf{2a - 2b - 4 - 5 + a}$

Ans: Simplify the brackets and combine the like terms.

$ \Rightarrow 2a - 2b - 4 - 5 + a = 2a + a - 2b - 4 - 5  \\ $

$ \Rightarrow 2a - 2b - 4 - 5 + a = 3a - 2b - 9 \\  $

 Substitute $ - 2$ for $b$ and $ - 1$ for $a$ and then solve.

$ \Rightarrow 2a - 2b - 4 - 5 + a = 3\left( { - 1} \right) - 2\left( { - 2} \right) - 9 \\ $

$ \Rightarrow 2a - 2b - 4 - 5 + a =  - 3 + 4 - 9 \\ $

$ \Rightarrow 2a - 2b - 4 - 5 + a = 1 - 9 \\ $

$ \Rightarrow 2a - 2b - 4 - 5 + a =  - 8 \\  $

The determined value is $ - 8$.

8. (i) If $\mathbf{z = 10}$, find the value of $\mathbf{{z^3} - 3\left( {z - 10} \right)}$.

Ans: Substitute the value of $z$ as $10$ then find a cube of $10$ to solve.

$ \Rightarrow {z^3} - 3\left( {z - 10} \right) = {\left( {10} \right)^3} - 3\left( {10 - 10} \right) \\ $

$ \Rightarrow {z^3} - 3\left( {z - 10} \right) = 1000 - 3 \times 0 \\ $

$ \Rightarrow {z^3} - 3\left( {z - 10} \right) = 1000 \\ $

The value of ${z^3} - 3\left( {z - 10} \right)$ is $1000$.

8. (ii) If $\mathbf{p =  - 10}$, find the value of $\mathbf{{p^2} - 2p - 100}$

Ans: Substitute the value of $p$ as $ - 10$ then find a square of $10$ to solve.

$  \Rightarrow {p^2} - 2p - 100 = {\left( {10} \right)^2} - 2\left( { - 10} \right) - 100  \\ $

$ \Rightarrow {p^2} - 2p - 100 = 100 + 20 - 100 \\ $

$ \Rightarrow {p^2} - 2p - 100 = 20  \\  $

The value of ${p^2} - 2p - 100$ is $20$.

9. What should be the value of $a$ if the value of $\mathbf{2{x^2} + x - a}$ equals to $5$, when $\mathbf{x = 0}$?

Ans: Substitute $0$ for $x$ in the expression $2{x^2} + x - a$ and equate it to $5$ in order to solve for $a$.

$ \Rightarrow 2{x^2} + x - a = 5 \\ $

$ \Rightarrow 2{\left( 0 \right)^2} + 0 - a = 5 \\ $

$ \Rightarrow 0 + 0 - a = 5 \\ $

$ \Rightarrow  - a = 5 \\ $

$ \Rightarrow a =  - 5 \\ $

So, the value of $a$ is $ - 5$.

10. Simplify the expression and find its value when $\mathbf{a = 5}$ and $\mathbf{b =  - 3}$:

$\mathbf{2\left( {{a^2} + ab} \right) + 3 - ab}$

Ans: Solve the brackets and combine the like terms with the variable $ab$. 

$ \Rightarrow 2\left( {{a^2} + ab} \right) + 3 - ab = 2{a^2} + 2ab + 3 - ab  \\ $

$ \Rightarrow 2\left( {{a^2} + ab} \right) + 3 - ab = 2{a^2} + 2ab - ab + 3 \\ $

$ \Rightarrow 2\left( {{a^2} + ab} \right) + 3 - ab = 2{a^2} + ab + 3 \\  $

Now substitute $5$ for $a$ and $ - 3$ for $b$ to solve.

$ \Rightarrow 2\left( {{a^2} + ab} \right) + 3 - ab = 2{\left( 5 \right)^2} + \left( 5 \right)\left( { - 3} \right) + 3 \\ $

$ \Rightarrow 2\left( {{a^2} + ab} \right) + 3 - ab = 2 \times 25 - 15 + 3 \\ $

$ \Rightarrow 2\left( {{a^2} + ab} \right) + 3 - ab = 50 - 12 \\ $

$ \Rightarrow 2\left( {{a^2} + ab} \right) + 3 - ab = 38  \\ $

The value of expression $2\left( {{a^2} + ab} \right) + 3 - ab$ is $38$.

NCERT Solutions for Class 7 Maths Chapter 10 Algebraic Expressions Exercise 10.2

Opting for the NCERT solutions for Ex 10.2 Class 7 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 10.2 Class 7 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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